How to Make Strong Metals
With High Ductility?
Yan Beygelzimer
Donetsk Institute of Physics and Technology
National Academy of Sciences of Ukraine
1
Outline
1. How do we try to do it?
First way: fragmentation
Second way: consolidation
2. How do we try to explain and to
predict it?
Mechanics of the processing
Structure, ductility, strength
3. Final Thoughts
2
How do we try to do it?
•Severe plastic deformation (SPD)
•Twist Extrusion
•First way: Fragmentation
•Second way: Consolidation
3
Some of the SPD Techniques
High Pressure Torsion
3d Forging
Equal Channel
Angular Extrusion
Twist Extrusion
(Y.Beygelzimer, 1999)
4
The idea of TE:
Twist channel
5
The idea of TE:
Twist channel
Equivalent strain e1
The shape and the dimensions of
the work-piece do not change!
6
The idea of TE:
Equivalent strain e2
Twist channel
Repeated twist extrusion leads to grain refinement
7
Twist Extrusion: Two in One
Fragmentation
Twist
Extrusion
Consolidation
8
Fragmentation
•Grains refinement. UFG materials
•Breaking of a brittle particles
9
Why should we care about grain
refinement during severe plastic
deformation?
This is one of few effective techniques for
obtaining ultrafine-grained (UFG) materials
10
Grain refinement
Coarse
SPD
11
Grain refinement
UFG
SPD
12
Ultrafine-Grained Materials

What are they?
– Metals with grain size ~10-1000 nm
 Why are they appealing?
– Significantly improved properties
– Qualitatively different properties not seen
in conventional materials
13
High strength and ductility
of UFG materials
!
!
R.Z. Valiev, I.V. Alexandrov, Y.T. Zhu and T.C. Lowe, “Paradox of
Strength and Ductility in Metals Processed by Severe Plastic
Deformation,” J. Mater. Res. 17, 5-8 (2002).
14
Twist extrusion of the Ti
50 mm
Initial state
50 mm
After 3 passes
15
Twist extrusion of the Ti
16
SPD of the Ti
17
Plates for traumatology
are made of UFG Ti
18
Twist extrusion of the Al-Mg-Sc-Zr alloy
Chemical composition:
Al - 3 wt.%Mg - 0,3 wt.%Sc – 0,10 wt.%Zr
Initial grain size dav=100 µm
Standard direct extrusion:
T=280-300ºC
dav= 0.455 µm
dmin= 0.129µm
dmax=1.032 mm
Twist extrusion: T=280-300ºC
5 passes CW + 1 pass CCW
dav=0.325 µm
dmin= 0.077µm
dmax=0.671 mm
19
SPD of the Cu
Yield stress,
Tensile strength
MPa
MPa
Uniform
elongation,
%
As received
80
145
40
TE, 3 passes,
with BP
393
426
18
*
320
360
14
420
440
15
440
470
25
State
ECAE, 2
passes
* ECAE, 16
passes,
without BP
* ECAE, 16
passes,
with BP
*N. Krasil’nikov, R.Valiev
20
ECAE of the Cu
Wei Wei, Guang Chen, Jing Tao Wang, Guo Liang Chen
Journal of Advanced Materials 2005 (in press)
21
Three present-day ways to increase
strength and ductility of UFG materials
•Cryogenic processing to produce a bimodal grain
structure
•Formation of second-phase particles to modify the
propagation of shear bands
•Developing aging and annealing treatments that can be
applied to the UFG materials in the post-processed
condition
22
Breaking of a brittle particles
SPD
23
Breaking of a brittle particles
SPD
24
TE of secondary Al alloy
Al-88%; Si-9,5% …
1 pass
Initial state
σy=50МPа
δ=1%
σy=205МPа
δ=14%
25
Twist Extrusion of phosphorous Cu (P 9%)
Initial state
Treatment
Direct Extrusion, µ=4.5
Twist Extrusion
1 pass
YS, МPа
YS,MPa
d,%
360
4
420
11
Back pressure = 200 MPa, T= 623 K
26
Twist extrusion of the Al-Mg-Sc-ZR alloys
Chemical composition:
Al – 3 wt.%Mg - 0,3 wt.%Sc – 0,15 wt.%Zr
SEM, As-cast structure
Cross-section
0.5µm
SEM, As-deformed structure,
TE – 5 passes, Tdef=280-300ºC
Longitudinal section
0.5µm
27
Consolidation
SPD
28
Consolidation
SPD
29
Consolidation of nanostructural
Cu powder by Twist Extrusion
Back pressure = 200 MPa, T= 473 K
Initial powder, D=250 µm
Compression test
after 2 passes:
Yield stress = 450 MPa
Breaking strain = 28%
1 pass
2 passes
Tensile test
after 2 passes:
Yield stress = 200 MPa
Breaking strain = 15%
30
Consolidation of nanostructural
Cu powder by Twist Extrusion
State
Density,
%
Diameter of the
coherentscattering region
L, nm
powder
-
100
TE, 1 path
99.2
36
TE, 2 paths
99.6
55
31
Consolidation of amorphous Al86Ni6Co2Gd6
melt-spun ribbons by Twist Extrusion
0,8
500
0,6
450
0,4
400
0,2
Ribbons
Extrusions
350
0,0
300 350 400 450 500 550
Extrusion temperature (K)
Microhardness and the volume
fraction of the amorphous phase in the
compacted samples.
Intensity (a.u.)
2
Microhardness (kgf/mm )
Al86(Ni,Co)8Gd6
Volume density of crystals
1,0
550
573K
513 K
AQ
20
40
60
80 100 120 140
2 (deg)
X-ray diffraction patterns
32
Consolidation of the cutting of the
secondary Al alloy by Twist Extrusion
d=12 mm
Yield stress =180-220 МPа;
d= 20-24%
33
How do we try to explain
and to predict it?
•The questions
•Mechanics of the processing
•The Problem
•The Model
•Predictions
34
The questions
•Parameters of the process?
•Strength and Ductility of the materials?
•Structure of the materials?
35
Mechanics of metal flow
36
Stream lines
1
2
3
4
5
6
7
8
37
Kinematically-admissible velocity field
1. Volume constancy condition

divV  0
2. Boundary condition
 
V  V0
 
V n 0
38
Kinematically-admissible velocity field


V  rotA
- form function

 Ax  0.5V0 y;
0

 Ay  0.5V0 x;

V0 tg
A


xy   ( x, y, z ) P( x, y, z );
 z
R

 0

P- function which is varied
 0
39
Finding function P on the
experimental stream lines
Experimental stream line
Theoretical stream line
3
 ( z)
0
p ( z)
3
0
0
2
4
6
8
10 12 14 16 18 20 22 24 26 28 30 32 34
z
36
36
40
Metal Deformation under Twist Extrusion
We showed that most of the deformation achieved by Twist
Extrusion is Simple Shear at the ends of the twist channel
41
Equivalent strain for TE pass
The average equivalent strain during one pass
e=tan()
Distribution of the strain
500
400
Z Axis
300
200
100
Yield stress, Cu
0
-15
-10
-5
0
5
X Ax
is
10
15
-8
-6
-4
-2
0
2
4
6
8
xis
YA
42
The Problem with
Theoretical Model
One of the main problems faced by any
theoretical model is the need to capture the
multi-level character of plastic deformation
43
Metal Structure is Determined by the
Image of the Loading Process
Grain refinement
P4,T4,..
.
e2
P3,T3,..
.
P2,T2,..
.
P1,T1,..
.
e1
44
But … The Image of The Loading Process
depends on this structure
The reason is that the structure defines the mechanical
properties of the materials.
45
Macro-Micro Interdependency
Metal Structure
Image of the Loading
Process
46
The Problem
• We are trying to produce a given
ultrafine homogeneous structure
• In reality, however, the specimen
may respond with a number of bad
things: highly inhomogeneous
structure, deformation localization or
fracture.
• The reason why this happens is
precisely the “Interdependency”.
47
Our approach to capture the
interdependency
Internal parameters
Continuum model
FEM
Internal parameters will allow us to account for the
interdependency between the stress-strain state and
the structure.
48
Internal parameters serve as special
envoys representing the micro-level
processes at the macro-level

S
49
Model of the material
50
Constitutive equations of the
Mises’s model
f  ij 
f  ij   0, eij  
 ij
 s 
f  ij     

 3
2
1
   ikd ik
3
2
RVE
e  0

1
1


     ik  d ik   ik  d ik  
3
3



51
Porous body with structurally
inhomogeneous matrix
 s 
2


f







 s

ij 
f  ij  

 1 3
  
   
 3

2
2
e
2
RVE
2
e  0
 


  
  1   k 0   
     

Beygelzimer Y. et al., (1994) Engin. Fracture Mech., 48, N5
Beygelzimer Y. Proceedings of International Workshop on Modelling
52
of Metal Powder Forming Processes, Grenoble, France, July 21-23, 1997
Plausible reasoning
RVE
 s 
f  ij     

 3
2
2


 s

f  ij  

 1  
  
   
 3

2
2
2
53
Loading surfaces
=0



54
Loading surfaces of the cutting
of the secondary Al alloy
, 150
, M Пa
MPa
4
100
3
2
50
1
0
100
200
300
400
500
600
700
800
,M Пa
P,PMPa
: 1- 30%, 2- 20%, 3- 10%, 4- 3%.
55
Porous body with structurally
inhomogeneous matrix at <<1
 s

f  ij   6a    
  
 3

d

   6 3a 
d
s
2
,s (d)
2
2
a
56
Breaking of a brittle particles
  1 exp( )   2 1  exp( )
Beygelzimer Y., Shevelev A. On the Development of Fracture Models for
Metal Forming// Russian Metallurgy (Metally), Vol., N5, p. 452–456, 2003
57
The model of grain refinement
and viscous fracture
Beygelzimer Y. Grain refinement versus voids
accumulation during severe plastic deformations of
polycrystals: Mathematical simulation, Mechanics of
Materials, V. 37, N7, p. 753-767, 2005
58
Scale
Main characters of the Model
500 nm
• Accumulative Zone – a “spring”, the
part of crystals in which dislocation
charges accumulate during plastic
deformation; AZs emerge due to the
inhomogeneity of shear along the
sliding plane.
• Void – a bit of an emptiness
0
• Embryo – an embryo of high-angle
boundary (a partial disclination)
59
Pictures of Main Characters
60
The Birth of an Accumulative Zone
There are regions of polycrystals where dislocations get
plugged during plastic deformation. Such regions cause
bendings of the crystalline lattice.
61
The Birth of an Accumulative Zone
The model postulates that AZs
emerge in two places:
1. hurdles that exist in
polycrystals before
deformation
2. high-angle boundaries that
emerge during deformation.
62
Relaxations of Accumulative Zones in
coarse grained materials
There are different relaxation mechanisms for
accumulative zones. When talking about large
cold deformations, we will distinguish two main
mechanisms:
1. Emergence of high angle boundaries
(leading to grain refinement)
2. Emergence of voids (leading to fracture)
63
Relaxations of Accumulative Zones in
coarse grained materials
64
Boundary sliding
boundary
b1
1
b2
b1
b1
 0.99
b2
1.5
b1
 0.80
b2
f
b2
1
f1( s)
0.5
f2( s)
b1
f
f3( s)
0
0.5
1
b2
0
0.1
Prandtl model
0.2
0.3
0.4
0.5
ss
b1
65
Relaxations of Accumulative Zones in
ultrafine grained materials
Sliding plane
d
66
Relaxations of Accumulative Zones in
ultrafine grained materials
67
Relaxations of Accumulative Zones in
ultrafine grained materials
68
Equations of the Model
Classical Plasticity Theory:
G(,)=0
P(,;µ)=0
General
Constitutive
Proposed model with internal parameters capturing the structure:
G(,)=0
General
S µ)=0
P(,; S,;
Constitutive
K(S,, ,;µ)=0
Kinetic

69
Loading function
  s S 
* 
f  ij ; , S   6a    
  
 3

2
2
2
3
*
3
2
  C3 N d c v
70
Kinetic equations
 dN
 d  C1  C 2 C5 N b F S   C3  C 4 N
N  0 Nb  0

 dN b
S 0 0
 d  C 4 N  C5 N b

 dS C N
N n 1  0 Nb n1  Nb n 

 5 b
0
S n1  S n 
 d S  S 0
.  n 1   n

3
N  0 Nb  0
 d
2 d 3 v  C 

C
N
3
c
6
 d
S 0 0

Quasimonotonic
loading
Cyclic
loading
71
Prediction
72
Prediction (i)
As
 
N  0,
d  dc ,
:
Nb  0,
0
Ideal Plasticity Land
Metals don’t fracture and don’t harden under sufficiently
high pressure when equivalent strain is very large, i.e.,
metals become ideally plastic.
The reason is boundary sliding
73
Prediction (ii)
Ductility grows
for sufficiently large equivalent strain
The reason is boundary sliding
74
The term responsible for the birth of voids
d
p
*
     B 
d

3
2
 *    C3 N  dc3v
0.015
с
alpha
0.01
*
0.005
grows
αDuctility
* decrease
forfor
0
sufficiently
large 5strain.
strain.
0
sufficiently
large
strain
10

75
Prediction (iii)
P
Grain refinement intensity grows and
fracture decreases with the increase of
pressure in the center of deformation.
76
The reason is p grows with pressure increase
0.1
420
840 1120
p, MPa
1
3
 Nс   p  p 
S   D   0 


с


77
Correspondence with experimental results:
Influence of pressure on the microstructure of
molybdenum at hydroextrusion:
e=0.36
30mm
e=0.60
e=0.92
Pb
Pb = 0,1 MPa
Pb = 800 MPa
78
Correspondence with experimental results:
Effect of pressure on fragment size distribution
Hydroextrusion of molybdenum (e=0.6).
f,%
40
, GPa
20
l, 100 µm
79
Use in Twist Extrusion
Twist Extrusion based
on hydro-mechanical
extrusion
Twist Extrusion based on
mechanical extrusion with
backpressure
80
Model-based prediction for Twist Extrusion
- Рb  0.1 MPa
- Рb = 1000 MPa
81
Prediction (iv)
To get intense grain refinement, one needs to choose
deformation schemes with small value of ductility
and to perform deformation under pressure.
Grain refinement intensity is typically higher for
simple shear than for uniaxial elongation
82
The reason is simple shear has
smaller ductility (under the same
c
pressure).
1
3
 Nс   p  p 
S   D   0 


с
c(0), tension
under pressure
c(0), torsion

3p

c(0) is
the metal
ductility at p=0
с(0)
S

Ductility diagram, steel 08X18H10T
(Experiment data V.Kolmogorov, et al., 1986)
83
Ductility diagrams
for various metals
show that, as a rule,
ductility с(0) of
tension greater than
that of torsion.
(Experiment data
V.Ogorodnikov and
I.Sivak, 1999)
84
Prediction (v)
Quasi-monotone deformations provide higher grain
refinement intensity than cyclic deformations.
0.8
0.8
-1
1/mkm
S, S,mm
NN,,1/mkm^2
mm-2
0.6
0.4
0.6
0.4
0.2
0.2
0
0
0.1
0.2

0.3
strain
0.4
0.5
0
0
0.1
0.2
0.3

0.4
0.5
strain
85
Cyclic deformations provide higher ductility than
quasi-monotone deformations
0.003
porosity
porosity
0.002
C5
0.001
0
0
0.1
0.2
0.3

strain
0.4
0.5
86
In order to increase the intensity of grain
refinement under cyclic deformation, one has
to increase the amplitude of deformation.
For example, in Twist Extrusion we combine
clockwise and counter-clockwise dies.
87
clockwise
clockwise
A
88
counter-clockwise
clockwise
2A
89
To avoid cyclic deformation we combine Twist Extrusion
with Spread Extrusion
Twist Extrusion
Spread Extrusion
90
Prediction (vi)
Sufficiently high pressure in the center of
deformation prevents strain localization
P
P
Layers of the voids
91
Prediction (vii)
Grain size distribution
0,52
0.5
2
f,
mm
f,
1
f(d)/S
f(d),1/mkm
3
mm
1
0
0
5
10
d , mm
15
d, m km
Self-similar stage
11
0
0d
c
1
d , mSdm
2
3
When there are
fragments with size dc
92
Self-similarity of Experimentally
Obtained Distributions
1.5
f,4
1
f, mkm^(-1)
6
93%
=4.61
f d
fxM(d)
mm 1
1.07
46%
2
0.5
0.28
.
0
0
0.5
d , mm
1
1.5
d, m km
Cu, torsion
(V.Panin et al., 1985)
2
2.5
0
0
0.5
1
1.5
d/M(d)
d
2
2.5
3
d
93
Prediction (iv)
During the self-similar stage of grain refinement, the fragment
boundary mesh in the cross-section of the specimen represents
a fractal set with dimension , 1< <2.
S
1

d
 1
.
1< <2 - fractal dimension.
During the self-similar stage  is constant
94
s  S
s 
.
were
    1,
1
d

Hall-Petch
0   1
During the self-similar stage  is constant
95
When sufficiently many indivisible fragments of
size dc appear, the self-similarity of the
boundary mesh gets violated. In this case

1
d
s 
1
d
S
.
96
Simulation of the grain refinement
processes by Cellular Model
Initial grain
RVE initial state
Plastic deformationD
Outer stress
D0
RVE new state
Stress on each
element n
Extension DnDt
and rotation nDt
of each element
Deformation
continues. Rotation
moments appear
М
М
Moment stresses
relax through grain
splitting
Beygelzimer Y., et al., Philosophical Magazine A, 79, N10, (1999)
97
Limiting Grain size distribution
(Cellular Model)
Cu
10µm
Grain size, μm
98
Final Thoughts
99
What do we hope for?
One can substitute the classical plasticity model by the
above model in any FEM package to directly compute the
stress-strained state of a metal and its interdependence with
the structure
S

Continuum model
FEM
100
What do we hope for?
101
What do we have?
• The model is relatively new
• The limits are not entirely explored
• A long way toward good parameter estimation
… but there are grounds for hope
102